question 3\n4 pts\neach side of a square is increasing at a rate of 3 cm/s. at what rate is the area of the…

question 3\n4 pts\neach side of a square is increasing at a rate of 3 cm/s. at what rate is the area of the square increasing when the area of the square is 9 cm²?\n9 $\frac{cm^{2}}{s}$\nthe correct answer is not listed.\n3 $\frac{cm^{2}}{s}$\n27 $\frac{cm^{2}}{s}$\n18 $\frac{cm^{2}}{s}$

question 3\n4 pts\neach side of a square is increasing at a rate of 3 cm/s. at what rate is the area of the square increasing when the area of the square is 9 cm²?\n9 $\frac{cm^{2}}{s}$\nthe correct answer is not listed.\n3 $\frac{cm^{2}}{s}$\n27 $\frac{cm^{2}}{s}$\n18 $\frac{cm^{2}}{s}$

Answer

Explanation:

Step1: Find the side - length of the square

Given the area formula of a square $A = s^{2}$, when $A = 9\ cm^{2}$, we solve for $s$: $A=s^{2}=9$, so $s = 3\ cm$ (since $s>0$ as it represents a length).

Step2: Differentiate the area formula with respect to time

We have $A = s^{2}$. Differentiating both sides with respect to time $t$ using the chain - rule, we get $\frac{dA}{dt}=2s\frac{ds}{dt}$.

Step3: Substitute the known values

We know that $\frac{ds}{dt}=3\ cm/s$ and $s = 3\ cm$. Substitute these values into the equation $\frac{dA}{dt}=2s\frac{ds}{dt}$: $\frac{dA}{dt}=2\times3\times3$. $\frac{dA}{dt}=18\ cm^{2}/s$.

Answer:

E. $18\frac{cm^{2}}{s}$